The solution is
$$
\frac{\mu^2}{h^3}t = \frac{1}{(1 - e^2)^{3/2}}\left[2\arctan\left[\sqrt{\frac{1 - e}{1 + e}}\tan\frac{\nu}{2}\right] - \frac{e\sqrt{1 - e^2}\sin\nu}{1 + e\cos\nu}\right]
$$
The second term will work when simplified but I have a arctanh.

arctan(iz)=i*arctanh(z). The difference between the two ways of writing it is probably whether you take e>1 or e<1, since you have things like sqrt(1-e) floating around, which could be imaginary or not.